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2 years ago

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A carpenter made a wooden structure using one cube and one triangular prism. The diagram shows the structure. The carpenter is o

nly going to paint the surface area of the front side of the wooden structure. To the nearest hundredth, what is the surface area that will be painted?

1 answer:

agasfer [191]2 years ago

6 0

Answer:

Option B

$3.75\ in^2$

Step-by-step explanation:

we know that

The surface area of the front side is equal to the area of a square face of the cube plus the area of a right triangle of the triangular prism

so

$SA=(1.5)^2+\frac{1}{2}(3.5-1.5)(1.5)$

$SA=2.25+1.50=3.75\ in^2$

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Maya scored 32 out of 40 for a Math's test and 41 out of 55 for a science's test. For which test did she score a lower percentag

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Answer:

He score lower on the 41 out of 55

Step-by-step explanation:

32/40 = .80 or 80%

41/55 = 0.75 or 75%

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2 years ago

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I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit

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Answer:

Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

Step-by-step explanation:

Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

If the denominator of $c$ is also $2$ , then the range for its numerator (call it $p$ ) would be $3 < p < 4$ . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than $2$ .

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

$\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}$ .

$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

$\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}$ .

$\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}$ .

$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .

Apparently $\displaystyle c = \frac{16}{10} = \frac{8}{5}$ works. $c = 1.6$ while $\displaystyle \frac{1}{c} = \frac{5}{8} = 0.625$ .

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There are three choices of jellybeans - grape ,cherry and orange. if the probability of getting a grape is 3/10 and the probabil

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When looking at probabilities, two ideas are always true.

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2) The sum of all the probabilites is 1.

Idea #2 works here. (For example, think of how a die has six things and the probability of each is 1/6. So 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1.) Let G = the probability of grape, C = probability of cherry and O = the probability of orange. From Idea #2, G + C + O = 1. Since we know G and C, then

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2 years ago

2. In the product of (4a3 + 5a2 – 11a) and (-15 + 3a – 7a2), the co-efficient of a4 is …………………… .

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Answer:

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Step-by-step explanation:

We could do this the long way and expand the whole product or note that the $a^{4}$ terms arise from the product of the a³ and a terms and the product of the a² and a² terms, that is

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Summing gives

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with coefficient - 23

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2 years ago

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